3.1305 \(\int \cos ^{\frac {11}{2}}(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=361 \[ \frac {2 a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (9 a^2 (9 A+11 C)+143 a b B+24 A b^2\right )}{693 d}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^3 B+3 a^2 b (7 A+9 C)+27 a b^2 B+3 b^3 (3 A+5 C)\right )}{15 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (77 a^3 B+33 a^2 b (7 A+9 C)+242 a b^2 B+24 A b^3\right )}{495 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d}+\frac {2 (11 a B+6 A b) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)^2}{99 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)^3}{11 d} \]

[Out]

2/15*(7*a^3*B+27*a*b^2*B+3*b^3*(3*A+5*C)+3*a^2*b*(7*A+9*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/231*(165*a^2*b*B+77*b^3*B+33*a*b^2*(5*A+7*C)+5*a^3*(9*A+11*C))*(cos(1/
2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/495*(24*A*b^3+77*a^3*B+242*
a*b^2*B+33*a^2*b*(7*A+9*C))*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/693*a*(24*A*b^2+143*a*b*B+9*a^2*(9*A+11*C))*cos(d*
x+c)^(5/2)*sin(d*x+c)/d+2/99*(6*A*b+11*B*a)*cos(d*x+c)^(3/2)*(b+a*cos(d*x+c))^2*sin(d*x+c)/d+2/11*A*cos(d*x+c)
^(3/2)*(b+a*cos(d*x+c))^3*sin(d*x+c)/d+2/231*(165*a^2*b*B+77*b^3*B+33*a*b^2*(5*A+7*C)+5*a^3*(9*A+11*C))*sin(d*
x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.96, antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4112, 3049, 3033, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (7 A+9 C)+7 a^3 B+27 a b^2 B+3 b^3 (3 A+5 C)\right )}{15 d}+\frac {2 a \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (9 a^2 (9 A+11 C)+143 a b B+24 A b^2\right )}{693 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (33 a^2 b (7 A+9 C)+77 a^3 B+242 a b^2 B+24 A b^3\right )}{495 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )}{231 d}+\frac {2 (11 a B+6 A b) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)^2}{99 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)^3}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(11/2)*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*(7*a^3*B + 27*a*b^2*B + 3*b^3*(3*A + 5*C) + 3*a^2*b*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(16
5*a^2*b*B + 77*b^3*B + 33*a*b^2*(5*A + 7*C) + 5*a^3*(9*A + 11*C))*EllipticF[(c + d*x)/2, 2])/(231*d) + (2*(165
*a^2*b*B + 77*b^3*B + 33*a*b^2*(5*A + 7*C) + 5*a^3*(9*A + 11*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (2
*(24*A*b^3 + 77*a^3*B + 242*a*b^2*B + 33*a^2*b*(7*A + 9*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(495*d) + (2*a*(2
4*A*b^2 + 143*a*b*B + 9*a^2*(9*A + 11*C))*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(693*d) + (2*(6*A*b + 11*a*B)*Cos[c
+ d*x]^(3/2)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(99*d) + (2*A*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^3*Sin
[c + d*x])/(11*d)

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 4112

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \cos ^{\frac {11}{2}}(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^3 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {2}{11} \int \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2 \left (\frac {1}{2} b (3 A+11 C)+\frac {1}{2} (9 a A+11 b B+11 a C) \cos (c+d x)+\frac {1}{2} (6 A b+11 a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 (6 A b+11 a B) \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {4}{99} \int \sqrt {\cos (c+d x)} (b+a \cos (c+d x)) \left (\frac {3}{4} b (15 A b+11 a B+33 b C)+\frac {1}{4} \left (150 a A b+77 a^2 B+99 b^2 B+198 a b C\right ) \cos (c+d x)+\frac {1}{4} \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 (6 A b+11 a B) \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {8}{693} \int \sqrt {\cos (c+d x)} \left (\frac {21}{8} b^2 (15 A b+11 a B+33 b C)+\frac {9}{8} \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \cos (c+d x)+\frac {7}{8} \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{495 d}+\frac {2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 (6 A b+11 a B) \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {16 \int \sqrt {\cos (c+d x)} \left (\frac {231}{16} \left (7 a^3 B+27 a b^2 B+3 b^3 (3 A+5 C)+3 a^2 b (7 A+9 C)\right )+\frac {45}{16} \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \cos (c+d x)\right ) \, dx}{3465}\\ &=\frac {2 \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{495 d}+\frac {2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 (6 A b+11 a B) \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {1}{15} \left (7 a^3 B+27 a b^2 B+3 b^3 (3 A+5 C)+3 a^2 b (7 A+9 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{77} \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (7 a^3 B+27 a b^2 B+3 b^3 (3 A+5 C)+3 a^2 b (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{495 d}+\frac {2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 (6 A b+11 a B) \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac {1}{231} \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (7 a^3 B+27 a b^2 B+3 b^3 (3 A+5 C)+3 a^2 b (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (165 a^2 b B+77 b^3 B+33 a b^2 (5 A+7 C)+5 a^3 (9 A+11 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (24 A b^3+77 a^3 B+242 a b^2 B+33 a^2 b (7 A+9 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{495 d}+\frac {2 a \left (24 A b^2+143 a b B+9 a^2 (9 A+11 C)\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 (6 A b+11 a B) \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{99 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3 \sin (c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]  time = 1.87, size = 286, normalized size = 0.79 \[ \frac {10 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 (9 A+11 C)+165 a^2 b B+33 a b^2 (5 A+7 C)+77 b^3 B\right )+154 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^3 B+3 a^2 b (7 A+9 C)+27 a b^2 B+3 b^3 (3 A+5 C)\right )+\frac {1}{12} \sin (c+d x) \sqrt {\cos (c+d x)} \left (154 \cos (c+d x) \left (43 a^3 B+3 a^2 b (43 A+36 C)+108 a b^2 B+36 A b^3\right )+5 \left (36 a \cos (2 (c+d x)) \left (a^2 (16 A+11 C)+33 a b B+33 A b^2\right )+154 a^2 (a B+3 A b) \cos (3 (c+d x))+3 \left (21 a^3 A \cos (4 (c+d x))+a^3 (531 A+572 C)+1716 a^2 b B+132 a b^2 (13 A+14 C)+616 b^3 B\right )\right )\right )}{1155 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(11/2)*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(154*(7*a^3*B + 27*a*b^2*B + 3*b^3*(3*A + 5*C) + 3*a^2*b*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2] + 10*(165*a^2*
b*B + 77*b^3*B + 33*a*b^2*(5*A + 7*C) + 5*a^3*(9*A + 11*C))*EllipticF[(c + d*x)/2, 2] + (Sqrt[Cos[c + d*x]]*(1
54*(36*A*b^3 + 43*a^3*B + 108*a*b^2*B + 3*a^2*b*(43*A + 36*C))*Cos[c + d*x] + 5*(36*a*(33*A*b^2 + 33*a*b*B + a
^2*(16*A + 11*C))*Cos[2*(c + d*x)] + 154*a^2*(3*A*b + a*B)*Cos[3*(c + d*x)] + 3*(1716*a^2*b*B + 616*b^3*B + 13
2*a*b^2*(13*A + 14*C) + a^3*(531*A + 572*C) + 21*a^3*A*Cos[4*(c + d*x)])))*Sin[c + d*x])/12)/(1155*d)

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{3} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{5} + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{4} + A a^{3} \cos \left (d x + c\right )^{5} + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b^3*cos(d*x + c)^5*sec(d*x + c)^5 + (3*C*a*b^2 + B*b^3)*cos(d*x + c)^5*sec(d*x + c)^4 + A*a^3*cos(
d*x + c)^5 + (3*C*a^2*b + 3*B*a*b^2 + A*b^3)*cos(d*x + c)^5*sec(d*x + c)^3 + (C*a^3 + 3*B*a^2*b + 3*A*a*b^2)*c
os(d*x + c)^5*sec(d*x + c)^2 + (B*a^3 + 3*A*a^2*b)*cos(d*x + c)^5*sec(d*x + c))*sqrt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3*cos(d*x + c)^(11/2), x)

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maple [B]  time = 6.37, size = 1082, normalized size = 3.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(11/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(20160*A*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-50400*A*a^3-36960*A*a^2*b-12320*B*a^3)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(56880*A*a^3+73920*A
*a^2*b+23760*A*a*b^2+24640*B*a^3+23760*B*a^2*b+7920*C*a^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-34920*A*a
^3-68376*A*a^2*b-35640*A*a*b^2-5544*A*b^3-22792*B*a^3-35640*B*a^2*b-16632*B*a*b^2-11880*C*a^3-16632*C*a^2*b)*s
in(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(13860*A*a^3+31416*A*a^2*b+27720*A*a*b^2+5544*A*b^3+10472*B*a^3+27720*B
*a^2*b+16632*B*a*b^2+4620*B*b^3+9240*C*a^3+16632*C*a^2*b+13860*C*a*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c
)+(-2790*A*a^3-5544*A*a^2*b-7920*A*a*b^2-1386*A*b^3-1848*B*a^3-7920*B*a^2*b-4158*B*a*b^2-2310*B*b^3-2640*C*a^3
-4158*C*a^2*b-6930*C*a*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+675*A*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2475*A*a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4851*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-2079*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3+2475*a^2*b*B*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1155*b^3*B*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1617*B*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-6237*B*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+825*C*(2*sin(1/2*d*x+1/2*c)
^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+3465*C*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-6237*C*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-3465*C*(2*sin(1/2*d*x
+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3)/(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 6.20, size = 514, normalized size = 1.42 \[ \frac {2\,\left (C\,b^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+C\,a\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+C\,a\,b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {B\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a\,b^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,A\,a^2\,b\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,B\,a\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,C\,a^2\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(11/2)*(a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*(C*b^3*ellipticE(c/2 + (d*x)/2, 2) + C*a*b^2*ellipticF(c/2 + (d*x)/2, 2) + C*a*b^2*cos(c + d*x)^(1/2)*sin(c
+ d*x)))/d + (B*b^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d - (2*A*a^3
*cos(c + d*x)^(13/2)*sin(c + d*x)*hypergeom([1/2, 13/4], 17/4, cos(c + d*x)^2))/(13*d*(sin(c + d*x)^2)^(1/2))
- (2*A*b^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^
(1/2)) - (2*B*a^3*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c
+ d*x)^2)^(1/2)) - (2*C*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*
(sin(c + d*x)^2)^(1/2)) - (2*A*a*b^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^
2))/(3*d*(sin(c + d*x)^2)^(1/2)) - (6*A*a^2*b*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, co
s(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)) - (6*B*a*b^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4]
, 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (2*B*a^2*b*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([
1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d*(sin(c + d*x)^2)^(1/2)) - (6*C*a^2*b*cos(c + d*x)^(7/2)*sin(c + d*x)*hy
pergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(11/2)*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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